Two dice are thrown simultaneously.

[given]

So, that number of possible outcomes = 36

(i) Sum of the numbers appearing on the dice is 7.

So, the possible ways are (1, 6), (2,5), (3, 4), (4, 3), (5, 2) and (6, 1).

Number of possible ways = 6

∴ Required probability = 6/36 = 1/6

(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.

So, the possible ways are (1, 1), (1,2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).

Number of possible ways = 15

∴ Required probability = 15/36 = 5/12

(iii) Sum of the numbers appearing on the dice is 1.

It is not possible, so its probability is zero.